[R] Regular expression help

[Eric Berger]

Hi Duncan,
You can try this:

library(readr)
f <- function(s) {
  t <- unlist(readr::tokenize(paste0(gsub(" ",",",s),"\n",collapse="")))
  i <- grep("[a-zA-Z0-9]*/[a-zA-Z0-9]*/",t)
  u <- sub("[a-zA-Z0-9]*/[a-zA-Z0-9]*/","",t[i])
  paste0(u,collapse=" ")
}

f("f 147/1315/587 2820/1320/587 3624/1321/587 1852/1322/587")
# "587 587 587 587"

f("f 1067 28680 24462")
# ""

HTH,
Eric


On Mon, Oct 9, 2017 at 6:23 PM, Ulrik Stervbo <ulrik.stervbo at gmail.com>
wrote:

> Hi Duncan,
>
> why not split on / and take the correct elements? It is not as elegant as
> regex but could do the trick.
>
> Best,
> Ulrik
>
> On Mon, 9 Oct 2017 at 17:03 Duncan Murdoch <murdoch.duncan at gmail.com>
> wrote:
>
> > I have a file containing "words" like
> >
> >
> > a
> >
> > a/b
> >
> > a/b/c
> >
> > where there may be multiple words on a line (separated by spaces).  The
> > a, b, and c strings can contain non-space, non-slash characters. I'd
> > like to use gsub() to extract the c strings (which should be empty if
> > there are none).
> >
> > A real example is
> >
> > "f 147/1315/587 2820/1320/587 3624/1321/587 1852/1322/587"
> >
> > which I'd like to transform to
> >
> > " 587 587 587 587"
> >
> > Another real example is
> >
> > "f 1067 28680 24462"
> >
> > which should transform to "   ".
> >
> > I've tried a few different regexprs, but am unable to find a way to say
> > "transform words by deleting everything up to and including the 2nd
> > slash" when there might be zero, one or two slashes.  Any suggestions?
> >
> > Duncan Murdoch
> >
> > ______________________________________________
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> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
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>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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